OK, this appears to me like perhaps a dumb question. I am reading Allen Hatcher's Algebraic Topology. I've seen bits and pieces of further material here and there before, now I'm restarting from the beginning.

OK, visually I can see why, say, the 2-sphere $S^2$ can not deformation retract onto it's equator. Intuitively, we can't do this without tearing a hole (or rather two holes). Even with visualizing the 'cylinder' $S^2$ x I (well as a volume) for the function $F$ to $S^1$,

$F:S^2 \times I \rightarrow S^1$,

$F(x,t) = f_t(x)$ the family of homotopic maps. You can see that this mapping cylinder won't work in trying to identify points $F(x,1)$ to points in the circle.

The mapping cylinder for a map and spaces $f:X \rightarrow Y$, is the quotient space

$[(X \times I) \amalg Y]/((x,1) \sim f(x))$.

Or even $S^1$ to its equator, $S^0$ the union of two separate points: not "deformation-retractable". Again, looking at $S^1 \times I$ doesn't appear to have a cont. function to $\{x\} \cup \{y\}$ for distinct points $x,y$ in $S^1$. It doesn't seem to have any mapping cylinder as well.

Disclaimer: OBVIOUSLY a circle is NOT homotopic to a point. Just in case anyone gets any wrong ideas as to where I'm going with this. It's just a question on my part :) If anyone could guide me to the a better intuition or visualization to correct the error of what I'm seeing.

However, if we just pick one point $x_0$ from $S^1$, then the mapping cylinder looks just like a cone, with $F(x,0) = x$, $F(x_0, t) = x_0$ (which is a line going down the cone from the circle to the bottom apex), and $F(x,1) = x_0$. And it looks like the mapping cylinder shows how to continuously deform the circle to the point, even if the point is an element of the circle.

I am thinking this "homotopic picture" of what merely looks like, the circle def. retracting to a point, is somehow misleading me, in the sense that I am just missing something or looking at it wrong or etc...

Can anyone elucidate?

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    $\begingroup$ I've added a title. Please verify that it is indeed what you are asking. $\endgroup$ – Alex Becker May 31 '13 at 3:45
  • $\begingroup$ Yes! Thanks. like I said I know it doesn't, the fundamental groups tell us why. But that is more symbolic than "visual", I was seeking a visual answer. $\endgroup$ – user80309 May 31 '13 at 3:51

First of all, let's clarify one thing. A circle does retract onto a point, because a retract of a circle to a point on it is just a constant map $r : S^1 \to \{p\}$. What you're really asking about is the fact that a circle doesn't deformation retract onto a point.

A deformation retract would be a homotopy $F : S^1 \times I \to S^1$ taking the circle to one of its points, so to deformation retract a circle to a point you'd need to retract it to a point on the circle, via a series of maps $F_t$ that map the circle to itself. Your map seems to be shrinking the circle to a point, which doesn't work because you're moving the rest of the circle off of itself into the "empty space inside of it," which isn't allowed.

In other words, you're viewing the circle as being embedded in the plane, like you'd draw a circle on a piece of paper. But, topologically, the points "inside the circle" don't exist -- there's only the circle itself.

  • $\begingroup$ Oh I get that the hole is outside the $S^1$ space certainly, and the circle is always "inside the circle" via the homotopy maps. But as it gets smaller, like as you go down the cone, wouldn't the circle, still being inside itself, all continuously map to the point? $\endgroup$ – user80309 May 31 '13 at 3:57
  • $\begingroup$ You're saying something like $F_t$ be the map sending the circle to a circle of radius $1-t$ tangent to the original circle at $p$. But For $0 < t < 1$, that map does not send the circle to itself, so it's not a homotopy. Each $F_t$ has to map the unit circle to the unit circle. In some very vague sense, you can't "shrink" it, not even a little bit. $\endgroup$ – Daniel McLaury May 31 '13 at 4:00
  • $\begingroup$ Yea ok that helps too thanks! $\endgroup$ – user80309 May 31 '13 at 4:04
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    $\begingroup$ @user80309: Dear user, One way to think of it is that if you really want to identify the circle at time $t$ (which has radius $1 - t$) with the original circle at time $t = 0$ (which has radius $1$), you'll need to do a division by $1-t$ to renormalize the radius; but then this won't be defined when $t = 1$. So your picture of the mapping cone, which is correct, can't be converted into a deformation retract of the circle onto one of its points. Regards, $\endgroup$ – Matt E May 31 '13 at 4:56
  • $\begingroup$ Reviving an old thread sorry, but, is it correct to say that if you try to deformation retract $S^1$ to point $p$ then $p$ must be in $S^1$ or you aren't actually doing a deformation retraction? $\endgroup$ – Bob Feb 26 '16 at 23:51

If I understand correctly, you are asking why a circle does not deformation retract onto a point. Recall that a deformation retract of $S^1$ onto a point $x_0\in S$ is a function $f:S^1\times I\to S^1$ such that $f(x,0)=x$ for all $x\in S^1$ and $f(x,1)=x_0$ for all $x\in S^1$. You propose drawing a cone with one end (corresponding to $0$) equal to $S^1$ and the other (corresponding to $1$) equal to $x_0$, and defining $f(x,t)$ to be the point obtained by travelling a distance $t$ along the line segment connecting $x$ at one end to $x_0$ at the other. The issue is that for $t\ne 0,1$ and $x\ne x_0$, we will have $f(x,t)\notin S^1$, so this map does not actually have range in $S^1$. Essentially, the problem with your cone visualization is that part of the cone lies outside $S^1$.

  • $\begingroup$ OH the homotopy cone has to be (or would have to be) entirely "inside" $S^1$. OK that takes care of it, thanks! $\endgroup$ – user80309 May 31 '13 at 3:58

I think your confusing homotopy equivalent and homeomorphic. The fact that you can wrap a disc into a point just means that they are homotopy equivalent, but not that they are homeomorphic, for that you would need a bijection. Since the map you describe is clearly not bijective, a point and $\mathbb{S}^1$ are not homeomorphic.

  • $\begingroup$ No not the disc, its boundary the 1-sphere. $\endgroup$ – user80309 May 31 '13 at 4:01

I think I am not agree with Alex B. The picture of that cone we have in our mind is just intuitive. In the cylinder of S^1, we have the top most circle identified with a point on the circle. If u remove that point then we shall have an open subset of that cylinder. Which will be homeomorphic to the cone without the vertex. Now it you take the union of this cone without vertex with that extra point, you have the cone of circle.


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